Higher Mathematics 2nd Paper · Classes 11-12 · Chapter 1 · 1.1

Real Numbers and Subsets of Real Numbers.

7, 0, −3, 22/7, √2, π: every one of them is a real number, but they do not all belong to the same family. Once you know which set each one falls into, and which set sits inside which, the rest of the chapter gets easier. At the end we prove why √2 can never be a fraction.

1. From integers to primes

The integers are all the positive and negative whole numbers together with zero: ℤ = {..., −3, −2, −1, 0, 1, 2, 3, ...}. Those smaller than zero are the negative integers, ℤ_<0 = {..., −3, −2, −1}.

The positive integers are the natural numbers, ℕ = {1, 2, 3, ...}. They were invented for counting, so they are also called counting numbers; ℤ_>0 = ℕ. Notice that ℕ starts at 1. Add zero and you get the non-negative integers, ℤ_≥0 = {0, 1, 2, 3, ...} = {0} ∪ ℤ_>0.

A natural number other than 1 that is divisible only by itself and 1 is a prime. The set is P = {2, 3, 5, 7, 11, 13, 17, 19, 23, ...}. 1 itself is not prime: the definition leaves it out on purpose.

2. Rational and irrational

A number that can be written as p/q, with p, q ∈ ℤ and q ≠ 0, is rational: Q = {p/q : p, q ∈ ℤ and q ≠ 0}. With q = 1, p/q is an integer, so a rational number is either a fraction or an integer. Dividing one integer by another need not give an integer: 8 ÷ 4 = 2, but 4 ÷ 8 = 1/2.

A number that cannot be written as p/q (p, q ∈ ℤ, co-prime, q ≠ 0) is irrational: Q′ = {x : x ∈ ℝ and x ∉ Q}. Examples are √2, √3, √5, √7, √11, e and π.

Write them as decimals and the two families separate. A rational number either stops or settles into a block that repeats: 7/50 = 0.14 (terminating), 6/11 = 0.545454... = 0.5̇4̇ (recurring). An irrational number does neither: √2 = 1.414213562..., π = 3.141592654....

All the rationals and irrationals together make the set of real numbers, ℝ = Q ∪ Q′. Georg Cantor was the first to give an exact definition of the real numbers, in 1871.

The sets nest: ℕ ⊂ ℤ ⊂ Q ⊂ ℝ, and Q′ ⊂ ℝ, but Q′ sits beside Q, not inside it. ℝ is itself a subset of the complex numbers ℂ, so ℕ ⊂ ℤ ⊂ Q ⊂ ℝ ⊂ ℂ. Drop a number below and see which box it comes to rest in.

A number that lands in one box also belongs to every box around it. Now the other way round: sort a few numbers into three groups.

3. √2 on the number line

All the real numbers lie on one infinitely long straight line, the number line or real line. Each point on it is one rational or irrational number, and each real number has exactly one point. Right of the origin is positive, left is negative.

You can see that an irrational number has a point too, using √2. Take OA₁ = 1 and draw A₁P = 1 perpendicular to OX at A₁. The hypotenuse of the right isosceles triangle OA₁P is OP = √(1² + 1²) = √2. An arc with centre O and radius OP cuts OX at Q, so OQ = √2.

4. Recurring decimal to fraction

You can show by hand that a recurring decimal is rational. If x = 2.090909... then 100x = 209.090909.... Subtract one from the other and the whole tail after the point cancels: 99x = 207, so x = 207/99 = 23/11, a rational number.

When a block of 3 digits repeats, multiply by 1000. If x = 3.274274274... then 1000x = 3274.274274..., and subtracting gives 999x = 3271, so x = 3271/999. The power of 10 equals the length of the repeating block. See below what happens with the wrong power.

5. Why √2 is irrational

The proof is by contradiction. Suppose √2 = p/q, where p and q are positive integers with no common factor but 1. Squaring gives p² = 2q², so p² is even and therefore p is even; write p = 2m. Then 4m² = 2q², that is q² = 2m², so q is even too. But if both p and q are divisible by 2, the assumption breaks. Hence √2 is irrational. The same argument shows √5, √7 and √11 are irrational.

The proof for √3 follows the same pattern. Suppose √3 = p/q (p, q co-prime); then p² = 3q², so 3 | p², and since 3 is prime, 3 | p. Putting p = 3m gives 9m² = 3q², that is q² = 3m², so 3 | q. Both are divisible by 3, against co-primality; so √3 is irrational.

The general rule: if m is a positive integer that is not a perfect square, √m is irrational. For some n, n² < m < (n + 1)², so n < √m < n + 1 and √m is not an integer. If √m = p/q (q > 1, p and q co-prime), then mq = p²/q. But p² and q are co-prime with q > 1, so p²/q is not an integer, while mq is one. That contradiction makes √m irrational.

6. Check yourself

Three questions in the exam's style. The first is about the order of the sets.

In the second, watch which set is inside which.

In the last, read the decimals: does it stop, repeat, or neither?

The whole note on eight cards.

Source: Higher Mathematics 2nd Paper, Classes 11-12 (NCTB), Chapter 1 “Real Numbers and Inequalities”, section 1.1.